A conical tent is made by using a semi-circular piece of canvas of radius 8 feet.

Question

A conical tent is made by using a semi-circular piece of canvas of radius 8 feet. Find the height of the tent and the number of cubic feet of air inside.

By manipulating I have found a way to get to the solutions provided by the textbook (that is $h = 4\sqrt{3}$ and $V = \frac{64}{3}\sqrt{3}\pi$) but I do not understand them. Here is what I have done so far

Let A be the area of the semi-circular piece of canvas of radius 8 so that

$ A = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi 8^2 = 32\pi$

Using the formula for the lateral area of a cone we have

$\pi rs = A = 32\pi\implies s = \frac{32\pi}{8\pi} = 4$

This is where I am stuck because I have a slant height smaller than the radius. However if I keep pushing forward I get

$s^2 = h^2 + r^2 \implies h^2 = s^2 - r^2 = 4^2 - 8^2 \implies h = \sqrt{|-48|} = 4\sqrt{3}$

I am quite close to the solution, but I cannot find a way to set up the problem correctly.

Answer 1

Some hints:

• If you take any cone, cut in a straight line from the top to the base and lay it flat, the resulting shape will be the sector of a circle. What is the relationship between the slant height of the cone and the radius of the circular sector?
• In the formula for the lateral area of a cone, $r$ is the radius of the circular base of the cone, which is not the same as the radius of the circular sector.
• However, the arc length of the sector must be the same as the circumference of the base of the cone. How does the length of a circular arc relate to the angle subtended at the center of the circle?

Answer 2

Observe that the tent is a cone with side length $s=8$ and a circular base of radius $r= 4$. Thus, its height is

$$h=\sqrt{s^2-r^2} = \sqrt{48} = 4\sqrt 3$$

The corresponding volume is

$$v = \frac 13 \pi r^2h = \frac {64\pi}{3}\sqrt 3$$